Question

If the value of \[{{i}^{2}}=-1\], then calculate the value of \[3{{i}^{2}}+{{i}^{3}}-{{i}^{4}}\].
(a) \[-4-i\]
(b) \[-2-i\]
(c) \[2+i\]
(d) \[4+i\]
(e) \[6+2i\]

Answer 1

Hint: To solve this question we will assume variables a, b, c to \[3{{i}^{2}},{{i}^{3}}\] & \[-{{i}^{4}}\] and we will use the identity that \[{{i}^{2}}=-1\]. Finally we will add them up to get the result.

Complete step-by-step solution:
We are given that the value of \[{{i}^{2}}=-1\], then the value of \[{{i}^{3}}\], \[{{i}^{4}}\] can be calculated separately.
Firstly we will calculate the value of \[{{i}^{3}}\] and \[{{i}^{4}}\], thus proceed to calculate the value of \[3{{i}^{2}}+{{i}^{3}}-{{i}^{4}}\].
We have, \[{{i}^{2}}=-1\].
Then, Let \[a=3{{i}^{2}},b={{i}^{3}},c=-{{i}^{4}}\].
We have to calculate the value of a + b + c,
Because, \[{{i}^{2}}=-1\].
\[\begin{align}
  & \Rightarrow 3{{i}^{2}}=\left( 3 \right)\left( -1 \right) \\
 & \Rightarrow 3{{i}^{2}}=-3 \\
\end{align}\]
Therefore, $a = -3$ ------- (1)
Now consider b.
\[b={{i}^{3}}\]
We have, \[{{i}^{2}}=-1\].
Multiplying ‘i’ both sides of the above equation,
\[\Rightarrow {{i}^{3}}=-i\]
\[\Rightarrow b = -i \] ---------- (2)
Now compute, \[c=-{{i}^{4}}\].
We have, \[{{i}^{2}}=-1\].
Multiplying, \[{{i}^{2}}=-1\] on both sides we have,
\[\begin{align}
  & {{i}^{4}}=\left( -1 \right)\left( -1 \right) \\
 & \Rightarrow {{i}^{4}}=1 \\
\end{align}\]
Now, \[c=-{{i}^{4}}=-1\].
Hence, $c = -1$ –------ (3)
Now a + b + c, using (1), (2) & (3) we have,
\[a+b+c=-3-i-1=-4-i\], which is option (a).
Therefore, \[3{{i}^{2}}+{{i}^{3}}-{{i}^{4}}=-4-i\], option (a) is correct.

Note: Another way to solve this question can be directly. Substituting, \[{{i}^{2}}=-1\], \[{{i}^{4}}=1\] & \[{{i}^{3}}=-i\] to get the result, then answer would come as \[3{{i}^{2}}+{{i}^{3}}-{{i}^{4}}=3\left( -1 \right)+\left( -i \right)-1=-4-i\], option (a).